\newproblem{lay:5_4_13}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.4.13}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Define $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $T(\mathbf{x})=A\mathbf{x}$ with $A=\begin{pmatrix}0 & 1 \\ -3 & 4\end{pmatrix}$.
	Find a basis $\mathcal{B}$ for $\mathbb{R}^2$ with the property $[T]_{\mathcal{B}}$ is diagonal.
}{
  % Solution
	Let's diagonalize $A$
	\begin{center}
		$A=PDP^{-1}=\begin{pmatrix}0 & 1 \\ -3 & 4\end{pmatrix}=\begin{pmatrix}-0.3162& -0.7071 \\ -0.9487 & -0.7071\end{pmatrix}
		   \begin{pmatrix}3& 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}-0.3162& -0.7071 \\ -0.9487 & -0.7071\end{pmatrix}^{-1}$
	\end{center}
	If we construct the basis $\mathcal{B}=\{(-0.3162,-0.9487),(-0.7071,-0.7071)\}$ is a basis in which the matrix of $T$ relative to $\mathcal{B}$ is
	\begin{center}
		$[T]_{\mathcal{B}}=\begin{pmatrix}3& 0 \\ 0 & 1\end{pmatrix}$
	\end{center}
}
\useproblem{lay:5_4_13}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
